I put some bold symbols in your question because the quantities involved are vectors. Can I ask you to confirm 1) the bold is OK, and that 2) you really meant to have the /3? I thought UNOT just mapped antipodally, so there should be no -3 there.
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twistor59Jan 1 '13 at 17:29

This definition of the Universal NOT seems to differ from the one given in the related question: the OP's (current) definition maps states antipodally but introduces a "damping" of the Bloch vector.
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Juan Bermejo VegaJan 1 '13 at 17:45

1 Answer
1

Note As it has been said in the comments, this definition of Universal-NOT gate seems to differ from others discussed in other posts [1]. This answer uses the definition proposed by the OP, i.e.

$$\rho=\frac{1}{2}(I+\vec{b}\cdot\vec{\sigma}) \quad \longrightarrow \quad U(\rho)=\frac{1}{2}\left(I-\frac{1}{3}\vec{b}\cdot\vec{\sigma}\right)$$
Where I use the symbol $I$ for the identity matrix to avoid confusion with 1 and $\vec{b}\in \mathbb{R}^3$ denotes the Bloch vector.

We write the density matrices of the computational basis states explicitly:
$$ \rho_a=|a\rangle\langle a |= \frac{1}{2}(I+\vec{b}_a\cdot\vec{\sigma}),$$
where $a\in\{0,1\}$. Expanding this expression readily yields the vectors $b_a$:
$$\vec{b}_a=(0,0,\pm1).$$
Applying your definition of $U$ to these density operators, the action of the operator on basis states can be obtained directly:
$$ U(|0\rangle\langle 0|)= \frac{1}{2}(I- \frac{1}{3}\sigma_z)=\frac{1}{3}|0\rangle\langle 0| +\frac{2}{3}|1\rangle\langle 1 |$$
$$ U(|1\rangle\langle 1|)= \frac{1}{2}(I+ \frac{1}{3}\sigma_z)=\frac{2}{3}|0\rangle\langle 0| +\frac{1}{3}|1\rangle\langle 1 |$$
We can observe that, because the factor $1/3$ that "damps" the Bloch vector, pure basis states evolve into mixed states; notice that, intuitively, states get closer to the totally mixed state $I/2$ if you make the Bloch vector $\vec{b}$ go to zero.

You are welcome. If my reply answers all your questions then you could accept it. (I would normally not say this, but since there has not been any activity in your post for a very long time I guess your original problem has been solved.)
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Juan Bermejo VegaJul 30 '13 at 10:03

I mean, this is one of sites policies that you should keep not only for this question, but also whenever you ask something in physics.SE: it is nice to accept some answer that you like, find useful or solves completely (or partially) your original problem. If you are not satisfied with the answers you should try to say why to the people who try to help you.
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Juan Bermejo VegaJul 30 '13 at 10:07